A proton of mass mp has same energy as that of a photon of wavelength lambda . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

Let E represent the identical energy value shared by both particles: Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength: The calculated ratio maps to option (3). Combine the core formulas: lambdamatter = hsqrt(2mE) and lambdalight = hcmathrmE. Dividing them smoothly yields the standard non-relativistic scaling ratio. Class 12 Physics: Dual Nature of Radiation and Matter

A proton of mass mp has same energy as that of a photon of wavelength lambda . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is

Let E represent the identical energy value shared by both particles: Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength: The calculated ratio maps to option (3). Combine the core formulas: lambdamatter = hsqrt(2mE) and lambdalight = hcmathrmE. Dividing them smoothly yields the standard non-relativistic scaling ratio. Class 12 Physics: Dual Nature of Radiation and Matter

A proton of mass mp has same energy as that of a photon of wavelength lambda . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

Let E represent the identical energy value shared by both particles: Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength: The calculated ratio maps to option (3). Combine the core formulas: lambdamatter = hsqrt(2mE) and lambdalight = hcmathrmE. Dividing them smoothly yields the standard non-relativistic scaling ratio. Class 12 Physics: Dual Nature of Radiation and Matter

A proton of mass mp has same energy as that of a photon of wavelength lambda . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is

Let E represent the identical energy value shared by both particles: Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength: The calculated ratio maps to option (3). Combine the core formulas: lambdamatter = hsqrt(2mE) and lambdalight = hcmathrmE. Dividing them smoothly yields the standard non-relativistic scaling ratio. Class 12 Physics: Dual Nature of Radiation and Matter

A proton of mass mp has same energy as that of a photon of wavelength lambda . If the proton is moving at…

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More Dual Nature of Radiation and Matter Previous-Year Questions — Page 3

Q4 jee_main_2025_29_jan_evening Photoelectric Effect and Stopping Potential

In an experiment with photoelectric effect, the stopping potential:

A. $increases with increase in the wavelength of the incident light$
B. $increases with increase in the intensity of the incident light$
C. $is ( 1e) times the maximum kinetic energy of the emitted photoelectrons$
D. $decreases with increase in the intensity of the incident light$

Solution

Related Formula

K = hν - φ = eVₛ

where, $K$ $K_{\max}$ = maximum kinetic energy of photoelectrons $Vₛ$ $V_s$ = stopping potential $e$ $e$ = fundamental electronic charge

Core Logic

By definition, the stopping potential $Vₛ$ $V_s$ is the negative potential applied to stop the most energetic photoelectrons from reaching the collector electrode.

From Einstein's photoelectric equation:

eVₛ = K Vₛ = Ke

Thus, the stopping potential is exactly $(1)/(e)$ $\frac{1}{e}$ times the maximum kinetic energy of the emitted photoelectrons. It does not depend on the intensity of light.

Pattern Recognition

Remember the primary features of the photoelectric effect:

Stopping potential depends linearly on frequency, and inversely on wavelength.
Intensity changes current, but has zero effect on stopping potential.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q15 jee_main_2025_03_april_morning Photoelectric Effect and Work Function

The work function of a metal is

3~eV

. The color of the visible light that is required to cause emission of photoelectrons is:

A. Green
B. Blue
C. Red
D. Yellow

Solution

Related Formula

Einstein's Photoelectric Equation:

Kmax = hν - φ = (hc)/(λ) - φ

For emission to occur, the photon energy must exceed the work function:

Ephoton > φ λ < λthreshold = (hc)/(φ)

Useful shortcut: $hc \approx 1240~eV\cdot nm$ $hc \approx 1240\mathrm{~eV\cdot nm}$.

Core Logic

Let's find the threshold wavelength $λthreshold$ $\lambda_{\text{threshold}}$ for a work function of $φ = 3~eV$ $\phi = 3\mathrm{~eV}$:

λthreshold = 1240~eV\cdot nm3~eV \approx 413.3~nm

To cause photoelectric emission, the wavelength of the incident light must be shorter than this threshold:

λ < 413.3~nm

Step 1: Comparing Visible Spectrum Colors

Let's look at the standard approximate wavelength ranges for visible light:

Red: $620 - 750~nm$ $620 - 750\mathrm{~nm}$ (Energy $\approx 1.65 - 2.0~eV$ $\approx 1.65 - 2.0\mathrm{~eV}$)
Yellow: $570 - 590~nm$ $570 - 590\mathrm{~nm}$ (Energy $\approx 2.1 - 2.2~eV$ $\approx 2.1 - 2.2\mathrm{~eV}$)
Green: $495 - 570~nm$ $495 - 570\mathrm{~nm}$ (Energy $\approx 2.2 - 2.5~eV$ $\approx 2.2 - 2.5\mathrm{~eV}$)
Blue: $450 - 495~nm$ $450 - 495\mathrm{~nm}$ (At lower end, approaching violet down to $380~nm$ $380\mathrm{~nm}$; Energy $\approx 2.5 - 3.3~eV$ $\approx 2.5 - 3.3\mathrm{~eV}$)

Only Blue light contains wavelengths extending below $413.3~nm$ $413.3\mathrm{~nm}$ (high enough photon energy to surpass $3~eV$ $3\mathrm{~eV}$). Therefore, blue light is required to cause photoelectric emission from this metal.

Pattern Recognition

Remember standard photon energies of visible colors: Blue/Violet photons have higher energy (typically $> 2.8~eV$ $> 2.8\mathrm{~eV}$), while Red/Yellow photons have much lower energy ( $< 2.2~eV$ $< 2.2\mathrm{~eV}$). For a higher work function like $3~eV$ $3\mathrm{~eV}$, only highly energetic blue/violet light can succeed.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Q jee_main_2025_04_april_morning Radiation Pressure and Momentum

A small mirror of mass

m

is suspended by a massless thread of length

l

. Then the small angle through which the thread will be deflected when a short pulse of laser of energy

E

falls normal on the mirror (

c =

speed of light in vacuum and

g =

acceleration due to gravity)

A. $θ = 3E4mcsqrtgl$ $\theta = \frac{3E}{4mcsqrt{gl}}$
B. $θ = Emcsqrtgl$ $\theta = \frac{E}{mcsqrt{gl}}$
C. $θ = E2mcsqrtgl$ $\theta = \frac{E}{2mcsqrt{gl}}$
D. $θ = 2Emcsqrtgl$ $\theta = \frac{2E}{mcsqrt{gl}}$

Solution

Related Formula

Force due to a completely reflecting beam:

F = (2P)/(c) = (2)/(c)(dE)/(dt)

Change in momentum:

Δ p = mv = \int F dt = (2E)/(c)

Work-Energy Theorem or Conservation of Mechanical Energy for small deflections:

gl(2 ²(θ)/(2)) = (v²)/(2)

For small angles $(θ)/(2) \approx (θ)/(2)$ .

Core Logic

Assuming perfect normal reflection from the mirror surface, the pulse imparts a momentum impulse of $(2E)/(c)$ $\frac{2E}{c}$ to the mass. This provides an initial velocity $v$ $v$ to the mirror. The mirror then swings up to a maximum angle $θ$ $\theta$ where kinetic energy converts entirely to gravitational potential energy.

Deflection of suspended mirror via laser pulse diagram for Q2 - JEE Main 2025 Morning

Step 1: Calculate Initial Velocity

From momentum change:

m(v - 0) = (2E)/(c) v = (2E)/(mc)

Step 2: Relate to Angular Deflection

Using conservation of mechanical energy:

mgl(1 - θ) = (1)/(2)mv²

gl(2 ²(θ)/(2)) = (v²)/(2)

Since $θ$ $\theta$ is very small, $(θ)/(2) \approx (θ)/(2)$ :

gl(2((θ)/(2))²) = (v²)/(2) gl(θ²)/(2) = (v²)/(2) glθ² = v²

Step 3: Solve for Theta

Substitute $v = (2E)/(mc)$ $v = \frac{2E}{mc}$ into the expression:

glθ² = ((2E)/(mc))² = (4E²)/(m²c²)

θ² = (4E²)/(m²c²gl) θ = 2Emc\sqrt(gl)

Pattern Recognition

This is a standard ballistic pendulum problem where the impulse is delivered by radiation pressure. Perfect reflection means momentum transfer is double the incident momentum $(2\cdot (E)/(c))$ $\left(2\cdot \frac{E}{c}\right)$.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter Class 11 Physics: System of Particles and Rotational Motion

Q10 jee_main_2025_04_april_morning Photoelectric Effect and Intensity

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below

A. Both A and R are true but R is NOT the correct explanation of A
B. A is false but R is true
C. A is true but R is false
D. Both A and R are true and R is the correct explanation of A

Solution

Related Formula

Einstein's photoelectric equation:

eVₛ = h u - φ$$

eV_s = h u - \phi$$

where:

$Vₛ$ $V_s$ = stopping potential

u$ = frequency of light

$φ$ $\phi$ = work function
Intensity formula: $I = (n h$

u)/(A · t)$I = \frac{n h

u}{A \cdot t}$ (where $n$ $n$ is rate of photons).

Core Logic

Assertion Analysis: Stopping potential $Vₛ$ $V_s$ depends strictly linearly on frequency

u$ and work function $φ$ $\phi$. It is completely independent of the beam intensity. Therefore, Assertion A is false.

Reason Analysis: Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given

u > u₀$

u > u_0$. Thus, Reason R is true.

Pattern Recognition

Stopping Potential $rightarrow$ $\leftrightarrow$ Frequency/Energy characteristic. Photo-current / Emission Rate $rightarrow$ $\leftrightarrow$ Photon Intensity/Flux counts.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

A proton of mass mₚ$\mathrm{m_p}$ has same energy as that of a photon of wavelength λ$\lambda$ . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Dual Nature of Radiation and Matter Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Comparing Visible Spectrum Colors

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Initial Velocity

Step 2: Relate to Angular Deflection

Step 3: Solve for Theta

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Dual Nature of Radiation and Matter Questions — jee_main_2025_28_jan_morning

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A proton of mass $mₚ$ $\mathrm{m_p}$ has same energy as that of a photon of wavelength $λ$ $\lambda$ . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.